The Bayes formula is
$$E^T\left(\frac{V(T)}{P(T,T)}\right) = \frac{E^Q\left(\frac{dP^T}{dQ}\frac{V(T)}{P(T,T)}\right)}{E^Q\left(\frac{dP^T}{dQ} \right)}$$
I only write $P(T,T)$ to be explicit, it is trivially equal to $1$.
But then I also came across this formula in Brownian Motiona Calculus, page 167: $$E^T\left(\frac{V(T)}{P(T,T)}\right) = E^Q\left(\frac{dP^T}{dQ}\frac{V(T)}{P(T,T)}\right)$$
Similar relations are used in Arbitrage Theory in Continuous Time page 402 and 403.
Could someone please help me reconcile the two equations. Because in the second equation we seem to have ignored the denominator.
The Radon-Nikodym derivative is a probability density function, so $E^Q[\frac{dP^T}{dQ}]=1.$ The first expression, is the general one because it works also for conditional expectations. The second one doesn't but when $\mathscr{F}_0= \{ \emptyset, \Omega \}$, both formulas agree. The general formula is $$E^T\left(\frac{V(T)}{P(T,T)}\right) = \frac{E^Q \left[\left(\frac{dP^T}{dQ}\frac{V(T)}{P(T,T)}\right)|\mathscr{F}_t\right]}{E^Q \left[ \left(\frac{dP^T}{dQ} \right)|\mathscr{F}_t \right]}.$$